# Setting up integral in polar coordinate for surface area of paraboloid

$$z = 16 - 3x^2 - 3y^2$$ above plane z = 4

I tried turning z into $$z= 16 - 3r^2$$ but it doesnt seem to be correct

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The surface of your paraboloid $S$ is a union of "infinitesimal lampshades" of area $${\rm d}\omega=2\pi r{\rm d}s==2\pi r\sqrt{dr^2+dz^2} =2\pi r\sqrt{1+z'^2(r)}\ dr\qquad(0\leq r\leq 2)\ .$$ As $z'(r)=-6r$ we obtain the following value for the total area of $S$: $$\omega(S)=2\pi \int_0^2\sqrt{1+36 r^2} r\ dr= {2\pi\over108}(1+36 r^2)^{3/2}\biggr|_0^2={\pi\over54}\bigl(145^{3/2}-1\bigr)\ .$$